你如何从给定的N个数字中测试所有可能的加法组合,使它们加起来得到给定的最终数字?
一个简单的例子:
要添加的数字集:N ={1,5,22,15,0,…} 期望结果:12345
你如何从给定的N个数字中测试所有可能的加法组合,使它们加起来得到给定的最终数字?
一个简单的例子:
要添加的数字集:N ={1,5,22,15,0,…} 期望结果:12345
当前回答
@KeithBeller的回答略有变化的变量名称和一些评论。
public static void Main(string[] args)
{
List<int> input = new List<int>() { 3, 9, 8, 4, 5, 7, 10 };
int targetSum = 15;
SumUp(input, targetSum);
}
public static void SumUp(List<int> input, int targetSum)
{
SumUpRecursive(input, targetSum, new List<int>());
}
private static void SumUpRecursive(List<int> remaining, int targetSum, List<int> listToSum)
{
// Sum up partial
int sum = 0;
foreach (int x in listToSum)
sum += x;
//Check sum matched
if (sum == targetSum)
Console.WriteLine("sum(" + string.Join(",", listToSum.ToArray()) + ")=" + targetSum);
//Check sum passed
if (sum >= targetSum)
return;
//Iterate each input character
for (int i = 0; i < remaining.Count; i++)
{
//Build list of remaining items to iterate
List<int> newRemaining = new List<int>();
for (int j = i + 1; j < remaining.Count; j++)
newRemaining.Add(remaining[j]);
//Update partial list
List<int> newListToSum = new List<int>(listToSum);
int currentItem = remaining[i];
newListToSum.Add(currentItem);
SumUpRecursive(newRemaining, targetSum, newListToSum);
}
}'
其他回答
这是R中的一个解
subset_sum = function(numbers,target,partial=0){
if(any(is.na(partial))) return()
s = sum(partial)
if(s == target) print(sprintf("sum(%s)=%s",paste(partial[-1],collapse="+"),target))
if(s > target) return()
for( i in seq_along(numbers)){
n = numbers[i]
remaining = numbers[(i+1):length(numbers)]
subset_sum(remaining,target,c(partial,n))
}
}
Java解决方案的Swift 3转换(by @JeremyThompson)
protocol _IntType { }
extension Int: _IntType {}
extension Array where Element: _IntType {
func subsets(to: Int) -> [[Element]]? {
func sum_up_recursive(_ numbers: [Element], _ target: Int, _ partial: [Element], _ solution: inout [[Element]]) {
var sum: Int = 0
for x in partial {
sum += x as! Int
}
if sum == target {
solution.append(partial)
}
guard sum < target else {
return
}
for i in stride(from: 0, to: numbers.count, by: 1) {
var remaining = [Element]()
for j in stride(from: i + 1, to: numbers.count, by: 1) {
remaining.append(numbers[j])
}
var partial_rec = [Element](partial)
partial_rec.append(numbers[i])
sum_up_recursive(remaining, target, partial_rec, &solution)
}
}
var solutions = [[Element]]()
sum_up_recursive(self, to, [Element](), &solutions)
return solutions.count > 0 ? solutions : nil
}
}
用法:
let numbers = [3, 9, 8, 4, 5, 7, 10]
if let solution = numbers.subsets(to: 15) {
print(solution) // output: [[3, 8, 4], [3, 5, 7], [8, 7], [5, 10]]
} else {
print("not possible")
}
func sum(array : [Int]) -> Int{
var sum = 0
array.forEach { (item) in
sum = item + sum
}
return sum
}
func susetNumbers(array :[Int], target : Int, subsetArray: [Int],result : inout [[Int]]) -> [[Int]]{
let s = sum(array: subsetArray)
if(s == target){
print("sum\(subsetArray) = \(target)")
result.append(subsetArray)
}
for i in 0..<array.count{
let n = array[i]
let remaning = Array(array[(i+1)..<array.count])
susetNumbers(array: remaning, target: target, subsetArray: subsetArray + [n], result: &result)
}
return result
}
var resultArray = [[Int]]()
let newA = susetNumbers(array: [1,2,3,4,5], target: 5, subsetArray: [],result:&resultArray)
print(resultArray)
我不喜欢上面看到的Javascript解决方案。下面是我使用部分应用、闭包和递归构建的一个:
好的,我主要关心的是,如果组合数组能满足目标要求,希望这样你就能找到剩下的组合了
这里只需要设置目标并传递组合数组。
function main() {
const target = 10
const getPermutationThatSumT = setTarget(target)
const permutation = getPermutationThatSumT([1, 4, 2, 5, 6, 7])
console.log( permutation );
}
我提出的当前实现
function setTarget(target) {
let partial = [];
return function permute(input) {
let i, removed;
for (i = 0; i < input.length; i++) {
removed = input.splice(i, 1)[0];
partial.push(removed);
const sum = partial.reduce((a, b) => a + b)
if (sum === target) return partial.slice()
if (sum < target) permute(input)
input.splice(i, 0, removed);
partial.pop();
}
return null
};
}
Thank you.. ephemient
我已经将上述逻辑从python转换为php..
<?php
$data = array(array(2,3,5,10,15),array(4,6,23,15,12),array(23,34,12,1,5));
$maxsum = 25;
print_r(bestsum($data,$maxsum)); //function call
function bestsum($data,$maxsum)
{
$res = array_fill(0, $maxsum + 1, '0');
$res[0] = array(); //base case
foreach($data as $group)
{
$new_res = $res; //copy res
foreach($group as $ele)
{
for($i=0;$i<($maxsum-$ele+1);$i++)
{
if($res[$i] != 0)
{
$ele_index = $i+$ele;
$new_res[$ele_index] = $res[$i];
$new_res[$ele_index][] = $ele;
}
}
}
$res = $new_res;
}
for($i=$maxsum;$i>0;$i--)
{
if($res[$i]!=0)
{
return $res[$i];
break;
}
}
return array();
}
?>